Dissolving $120 \mathrm{~g}$ of a compound (mol. wt $=60$ ) in $1000 \mathrm{~g}$ of water gave a solution…

Dissolving $120 \mathrm{~g}$ of a compound (mol. wt $=60$ ) in $1000 \mathrm{~g}$ of water gave a solution of density $1.12 \mathrm{~g} \mathrm{~mL}^{-1}$. The molarity of solution is
  1. $1.0 \mathrm{M}$
  2. $2.0 \mathrm{M}$
  3. $2.5 \mathrm{M}$
  4. $4.0 \mathrm{M}$

Solution

Given, density $(d)=1.12 \mathrm{~g} \mathrm{~mL}^{-1}$ Mass of solute $(w)=120 \mathrm{~g}$ $\begin{aligned} & \text { Molar mass of solute }(M)=60 \\ & \text { Mass of solvent }(w)=1000 \mathrm{~g}=1000 \mathrm{~mL} \\ & \therefore \text { Total mass }=1000+120=1120 \mathrm{~g} \\ & \because \quad d=\frac{\text { Mass }}{\text { Volume }(V)} \\ & \therefore \text { Total volume }(V)=\frac{1120}{1.12}=1000 \mathrm{~mL} \\ & \because \text { Molarity }(c)=\frac{w}{M} \times \frac{1000}{V} \\ & \because \text { Molarity }(c)=\frac{w}{m} \times \frac{1000}{1000}=\frac{120}{60}=2.0 \mathrm{~m}\end{aligned}$

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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