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(B) Molarity $(M) = \frac{	ext{mass of solute (g)}}{	ext{molar mass (g/mol)}} 	imes \frac{1000}{	ext{volume of solution (mL)}}$.Mass of solution $

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(B) Molarity $(M) = \frac{	ext{mass of solute (g)}}{	ext{molar mass (g/mol)}} 	imes \frac{1000}{	ext{volume of solution (mL)}}$.Mass of solution $= 	ext{mass of solute} + 	ext{mass of solvent} = 50 \ g + 1000 \ g = 1050 \ g$.Volume of solution $(V) = \frac{	ext{mass of solution}}{	ext{density}}$.At $90^{\circ} C$,$V_1 = \frac{1050}{1.1} \ mL$,so $M_1 = \frac{50}{M_w} 	imes \frac{1000}{1050/1.1} = \frac{50 	imes 1000 	imes 1.1}{M_w 	imes 1050} = \frac{52.38}{M_w}$.At $10^{\circ} C$,$V_2 = \frac{1050}{1.15} \ mL$,so $M_2 = \frac{50}{M_w} 	imes \frac{1000}{1050/1.15} = \frac{50 	imes 1000 	imes 1.15}{M_w 	imes 1050} = \frac{54.76}{M_w}$.$\%$ change in molarity $= \frac{M_2 - M_1}{M_1} 	imes 100 = \frac{54.76/M_w - 52.38/M_w}{52.38/M_w} 	imes 100 = \frac{2.38}{52.38} 	imes 100 \approx 4.54 \% \approx 4.5 \%$.

$50 \ g$ of a substance is dissolved in $1 \ kg$ of water at $90^{\circ} C$. The temperature is reduced to $10^{\circ} C$. The density is increased from $1.1 \ g \ cm^{-3}$ to $1.15 \ g \ cm^{-3}$. What is the $\%$ change of molarity of the solution?

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