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#### 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.820C.  If Na2SO4 is 81.5% ionised, the value of x (K for water=1.860C kg mol−1) is approximately : (molar mass of S=32 g mol−1 and that of Na=23 g mol−1) Option: 1  15 g Option: 2  25 g Option: 3  45 g Option: 4  65 g

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#### The freezing point of benzene decreases by 0.450C when 0.2 g of acetic acid is added to 20 g of benzene.  If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (Kf for benzene=5.12 K kg mol−1) Option: 1  74.6% Option: 2 94.6 Option: 3  64.6% Option: 4 80.4%

In benzene ,

$2 CH_{3}COOH \rightleftharpoons \left ( CH_{3}COOH \right )_{2}$

t = 0                   1                                 -

t = t                      1 - $\beta$                          $\frac{\beta }{2}$

Given:

w = 0.2g                W = 20 g                  $\Delta$ T = 0.45 K

As we know , $\Delta T = \frac{1000 \times K_{f} \times w}{M \times W }$

$\Delta T = \frac{1000 \times 5.12 \times 0.2}{M \times 20 }= 0.45$

$\therefore$ , observed M = 113.78 (acetic acid)

Molecular weight of acetic acid = 60

$i =\frac{normal \; molecular \; mass}{observed \; molecular \; mass}$

$\therefore \frac{m_{normal}}{m_{observed}} = 1 - \beta + \frac{\beta }{2}$

$\frac{60}{113.78} = 1 - \beta + \frac{\beta }{2}$

$\Rightarrow \beta = 0.945$

% degree of association = 94.5 %

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#### At room temperature, a dilute solution of urea is prepared by dissolving 0.60g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, lowering of vapour pressure(in  mmHg)  will be: (molar mass of urea = 60g/mol) Option: 1 0.017 Option: 2 0.028 Option: 3 0.027 Option: 4 0.031

Molecular weight of urea = 60 g / mol

Now, we know that :

Relative lowering of vapour pressure is equal to the molar mass of the solute.

$\therefore \Delta P=35\times\frac{(\frac{0.6}{60})}{\frac{360}{18}+\frac{0.6}{60}}$

$\Delta P=35\times\frac{0.01}{20+0.01}$

$\Delta P=35\times0.00049$

$\Delta P=0.0174 \: mm Hg$

$\therefore$ option (1) is correct.

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#### The elevation of boiling point of 0.1 m aq $CrCl_{3}.xNH_{3}$ solution is two times that of 0.05 aq. $CaCl_{2}$  solution. The value of x _____ (Assume 100% ionization of complex and $CaCl_{2}$ , coordination number of Cr is 6 and that all of $NH_{3}$ molecules are present inside the coordination sphere.)

$\mathrm{\Delta T_{b}\; of \;CrCl_{3}.xNH_{3}= 2(\Delta T_{b}\; of \; CaCl_{2})}$

$i \times 0.1 = 2 \times 3 \times 0.05$

Thus, i = 3

Therefore, the complex is dissociated into 3 ions.

Hence, complex is $[Cr(NH_{3})_{5}Cl]Cl_{2}$

Thus, the value is 5.

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#### A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A,B and C. The relative lowring of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A=100 g mol -1; B=200 g mol -1; C=10,000 g mol -1]Option: 1Option: 2Option: 3Option: 4

Here vapour pressure of water is lowering.

We know this formula,

Partial Vapour pressure of water Pwater = PoXwater

Lowering Vapour pressure of water = Po – Pwater

The relative lowering of vapour pressure  $\mathrm{=\frac{P^o-P}{P^o}}=\mathrm{\frac{\Delta P}{P^o}} =1- \chi_{water}$

$\dpi{120} \mathrm{\mathrm{\frac{\Delta P}{P^o}} =\chi_{solute}=\frac{\frac{m_{solute}}{M_{solute}}}{\frac{m_{solute}}{M_{solute}}+\frac{m_{water}}{M_{water}}}}$

Now,

$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{A}}=\frac{\frac{10}{100}}{\frac{10}{100}+\frac{180}{18}}$

$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{B}}=\frac{\frac{10}{200}}{\frac{10}{200}+\frac{180}{18}}$

$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{C}}=\frac{\frac{10}{10,000}}{\frac{10}{10,000}+\frac{180}{18}}$

From above,

$\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{A}}>\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{B}}>\left(\frac{\Delta \mathrm{P}}{\mathrm{P}^{0}}\right)_{\mathrm{C}}$

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#### How much amount of NaCl (in grams) should be added to 600g of water to decrease the freezing point of water to $-0.2^{\circ}C$? (Assume that the freezing point depression constant for water = 2K kg mol-1) Option: 1 1.76 Option: 2 3.52 Option: 3 4.9 Option: 4 8.58

$\begin{array}{l}{\Delta \mathrm{T}_{\mathrm{f}}=0.2^{\circ} \mathrm{C}} \\ {\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik_{f}m}} \\ {0.2=2 \times 2 \times \frac{w}{58.5} \times \frac{1000}{600}} \\\\ {w=\frac{0.2 \times 58.5 \times 600}{1000 \times 4}} {=\frac{1.2 \times 58.5}{40}=1.76 \mathrm{g}}\end{array}$

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#### The osmotic pressure of a solution of NaCl is 0.10 atm and that of a glucose solution is 0.20 atm. The osmotic pressure of a solution formed by mixing 1L of the sodium chloride solution 2L of the glucose solution is . x is _________ (nearest interger)

We know this formula.

$\mathrm{ \pi_{f} v_{f}=\pi_{1} v_{1}+\pi_{2} v_{2} }$

$\mathrm{ \pi_{f} =\frac{\pi_{1} v_{1}+\pi_{2} v_{2}}{v_f} }$

$\mathrm{ \pi_{f} = \frac{0.1 \times 1+0.2 \times 2}{3} }$

$\mathrm{ \pi_{f} =\frac{0.5}{3}=\frac{500}{3} \times 10^{-3} }$

$\mathrm{ x =167 }$

Ans = 167

------------------------------------------

Alternate Solution

$\mathrm{ x =167 }$

Ans = 167

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#### The mole fraction of glucose $(C_{6}H_{12}O_{6})$  in an aqueous binary solution is 0.1. The mass percentage of water in it to the nearest integer is-

Mole fraction of glucose, in $C_{6}H_{12}O_{6}=0.1$

Now, lets assume we have 10 moles of solution.

Thus, moles of $C_{6}H_{12}O_{6}=1$

Ans, moles of $H_2O = 9$

Now, the mass of water = number of moles x molar mass

Thus, the mass of water  $\mathrm{= 9 \times 18 = 162 \ g}$

Mass of $C_{6}H_{12}O_{6}=180\: gm$

Now, mass percent of water is given as:

$\mathrm{Mass\: \%\: of\: water\: =\: \frac{Mass\: of\: water}{Mass\: of\: water\: +\: mass\: of\: glucose}\: x\: 100}$

$\mathrm{Mass\: \%\: of\: water\: =\: \frac{162}{180 +\: 162}\: x\: 100}$

$\mathrm{Mass\: \%\: of\: water\: =\: 47.3\%}$

Thus, the correct answer is 47.

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#### If $\inline 250\; cm^{3}$ of an aqueous solution containing $\inline 0.73\; g$ of a protein A is isotonic with one litre of another aqueous solution containing $\inline 1.65\; g$ of a protein $\inline B$, AT $\inline 298\; K$, the ratio of the molecular masses of $\inline A$ and $\inline B$ is ______$\inline \times 10^{-2}$ (to the nearest integer).

Let molar mass of protein A = x g/mol
Let molar mass of protein B = y g/mol

$\pi_{\mathrm{A}}=\text { osmotic pressure of protein } \mathrm{A}=\frac{\left(\frac{0.73}{\mathrm{x}}\right)}{0.25} \mathrm{RT}$

$\pi_{\mathrm{B}}=\text { osmotic pressure of protein } \mathrm{B}=\frac{\left(\frac{1.65}{\mathrm{y}}\right)}{1} \mathrm{RT}$

Now, $\pi_{\mathrm{A}}=\pi_{\mathrm{B}}$

$\left(\frac{0.73}{\mathrm{x} \times 0.25}\right) \mathrm{RT}=\left(\frac{1.65}{\mathrm{y}}\right) \mathrm{RT}$

$\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\frac{0.73}{0.25 \times 1.65}=1.769 \cong 1.77$

$\mathrm{\frac{x}{y} = 177 \times 10^{-2}}$

Ans = 177

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#### 2 molal solution of a weak acid HA has a freezing point of 3.885oC. The degree of dissociation of this acid is _________ (Round off to the Nearest Integer). [Given : Molal depression constant of water Freezing point of pure water = 0o C]

Given,

2 molal solution of a weak acid HA has a freezing point of 3.885ºC.

$\mathrm{T}_{\mathrm{f} \text { sol. }}=3.885^{\circ} \mathrm{C}$

m = 2

Molal depression constant of water (Kf) = 1.85 K kg mol-1

The freezing point of pure water  Ti = 0ºC

van't Hoff factor i is related to the degree of dissociation \alpha $\alpha$ as, if n = number of ions,

$\alpha=\frac{i-1}{n-1}$

Here n = 2 (H+and A-

So,

$i = 1+\alpha$

We know the formula of depression at the freezing point.

$\mathrm{\Delta T = i\times K_f\times m}$

$\mathrm{\Delta T = T_f -T_i= i\times K_f\times m}$

$3.885 - 0= i\times 1.85\times 2$

i = 1.05

Now,

$i = 1+\alpha$

$1.05 = 1+\alpha$

$\alpha=0.05$

$\alpha=50\times10^{-3}$

Ans = 50

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