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  • Solution of two components containing n1 moles of 1st component and n2 moles of 2nd component is prepared, M1 and M2 are molecular weight of component 1 and 2 respectively. If d is the density of the solution in  g mL-

Solution of two components containing n1 moles of 1st component and n2 moles of 2nd component is prepared, M1 and M2 are molecular weight of component 1 and 2 respectively. If d is the density of the solution in  g mL-1, C2 is the molarity and x2 is the mole fraction of 2nd component, C2 can be expressed as:
Option: 1 c_{2}=\frac{dx_{2}}{M_{2}+x_{2}(M_{2}-M_{1})}
 
Option: 2 c_{2}=\frac{dx_{2}}{M_{2}+x_{2}(M_{2}-M_{1})}
 
Option: 3 c_{2}=\frac{dx_{2}}{M_{2}+x_{2}(M_{2}-M_{1})}
 
Option: 4 c_{2}=\frac{dx_{2}}{M_{2}+x_{2}(M_{2}-M_{1})}
 
Option: 5 c_{2}=\frac{dx_{1}}{M_{2}+x_{2}(M_{2}-M_{1})}
Option: 6 c_{2}=\frac{dx_{1}}{M_{2}+x_{2}(M_{2}-M_{1})}
Option: 7 c_{2}=\frac{dx_{1}}{M_{2}+x_{2}(M_{2}-M_{1})}
Option: 8 c_{2}=\frac{dx_{1}}{M_{2}+x_{2}(M_{2}-M_{1})}
Option: 9 c_{2}=\frac{1000dx_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}  
Option: 10 c_{2}=\frac{1000dx_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}  
Option: 11 c_{2}=\frac{1000dx_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}  
Option: 12 c_{2}=\frac{1000dx_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}  
Option: 13 c_{2}=\frac{1000x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}
Option: 14 c_{2}=\frac{1000x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}
Option: 15 c_{2}=\frac{1000x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}
Option: 16 c_{2}=\frac{1000x_{2}}{M_{1}+x_{2}(M_{2}-M_{1})}

Answers (1)

best_answer

Molarity \(C_2\) is moles of solute per liter of solution.

Using mole fraction \(x_2\), molecular weights \(M_1, M_2\), and density \(d\), we derive: \[ C_2=\frac{1000 \cdot d \cdot x_2}{M_1+x_2\left(M_2-M_1\right)} \]

Posted by

Divya Sharma

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