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Chemistry Question on Stoichiometry and Stoichiometric Calculations

The molarity of urea (molar mass \ce60 g mol−1\ce{60 \, g \, mol^{-1}}) solution by dissolving 15 g of urea in 500 cm3cm^3 of water is

A

\ce2 mol dm−3\ce{2 \, mol \, dm^{-3}}

B

\ce0.5 mol dm−3\ce{0.5 \, mol \, dm^{-3}}

C

\ce0.125 mol dm−3\ce{0.125 \, mol \, dm^{-3}}

D

\ce0.0005 mol dm−3\ce{0.0005 \, mol \, dm^{-3}}

Answer

\ce0.5 mol dm−3\ce{0.5 \, mol \, dm^{-3}}

Explanation

Solution

Given, mass of urea =15 g=15 \,g

Molar mass of urea =60 g mor−1=60 \,g \,mor ^{-1}

Volume =500 cm3=500 mL=0.5 L=500\, cm ^{3}=500\, mL =0.5 \,L

[∵1 cm3=1 mL 1 L=1000 mL]\begin{bmatrix}\because1\,cm^{3}=1\,mL\\\ 1\,L=1000\,mL\end{bmatrix}

∴\therefore Molarity

= Mass of solute in grams (W) Molar mass of solute (M)× Volume of solution V(L)=\frac{\text { Mass of solute in grams }(W)}{\text { Molar mass of solute }(M) \times \text { Volume of solution } V(L)}

Molarity of urea =15 g60 g mor−1×0.5 L=\frac{15\, g }{60 \,g\, mor ^{-1} \times 0.5\, L }

⇒0.5 mol/L=0.5mol/dm3\Rightarrow 0.5 \,mol / L =0.5 mol / dm ^{3}

(∵1L=1 dm3)\left(\because 1 L =1 \,dm ^{3}\right)