100 g of solute is dissolved in 1400 g of s | vedclass

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(C) Let the mass of solute be $W_B = 100 \ g$ and mass of solvent be $W_A = 1400 \ g$.Total mass of solution $= W_A + W_B = 1400 + 100 = 1500 \ g$.Den

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(C) Let the mass of solute be $W_B = 100 \ g$ and mass of solvent be $W_A = 1400 \ g$.Total mass of solution $= W_A + W_B = 1400 + 100 = 1500 \ g$.Density of solution $d = 1.5 \ g/mL$.Volume of solution $V = \frac{	ext{mass}}{d} = \frac{1500 \ g}{1.5 \ g/mL} = 1000 \ mL = 1 \ L$.Molarity $(M) = \frac{n_B}{V(L)} = \frac{W_B / M_B}{1} = \frac{100}{M_B}$.Molality $(m) = \frac{n_B}{W_A(kg)} = \frac{100 / M_B}{1400 / 1000} = \frac{100}{M_B} 	imes \frac{1000}{1400} = \frac{100}{M_B} 	imes \frac{10}{14}$.Ratio $\frac{M}{m} = \frac{100 / M_B}{(100 / M_B) 	imes (10 / 14)} = \frac{14}{10} = 1.4$.

$100 \ g$ of solute is dissolved in $1400 \ g$ of solvent. The density of the resultant solution is $1.5 \ g/mL$. The ratio of its molarity $(M)$ to molality $(m)$ is:

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