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Two solutions \mathrm{A} and \mathrm{B} are prepared by dissolving 1 \mathrm{~g} of non-volatile solutes \mathrm{X} and \mathrm{Y, } respectively in \mathrm{1 \mathrm{~kg}} of water. The ratio of depression in freezing points for \mathrm{A} and \mathrm{\mathrm{B}} is found to be \mathrm{1: 4. }The ratio of molar masses of \mathrm{ \mathrm{X}} and \mathrm{\mathrm{Y}} is

Option: 1

1: 4


Option: 2

1: 0.25


Option: 3

1: 0.20


Option: 4

1: 5


Answers (1)

best_answer

We know,

\mathrm{\Delta T_f \propto m} , if solvent is same

\mathrm{\frac{\left(\Delta T_f\right)_1}{\left(\Delta T_{\left.f_2\right)_2}\right.}=\frac{m_1}{m_2}=\frac{\omega_{x / M_x}}{\omega_{\text {water }}} \times \frac{w_{\text {vater }}}{\omega_y / M_y}}

\mathrm{\frac{1}{4}=\frac{\omega_x x}{M_x} \times \frac{M_y}{\omega_y}}

\mathrm{\frac{1}{4}=\frac{M_y}{M_x} \quad\left(\because \omega_x=1 \mathrm{~g} \text { and } \omega_y=1 \mathrm{~g}\right)}

\mathrm{\frac{M_x}{M_y}=\frac{4}{1}=\frac{1}{1 / 4}=\frac{1}{0.25}}

Hence, the correct answer is Option (2).

Posted by

Rakesh

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