Question:

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

Updated On: Jan 23, 2024
  • $\ce{2 \, mol \, dm^{-3}}$
  • $\ce{0.5 \, mol \, dm^{-3}}$
  • $\ce{0.125 \, mol \, dm^{-3}}$
  • $\ce{0.0005 \, mol \, dm^{-3}}$
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The Correct Option is B

Solution and Explanation

Given, mass of urea $=15 \,g$

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

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

$\begin{bmatrix}\because1\,cm^{3}=1\,mL\\ 1\,L=1000\,mL\end{bmatrix}$

$\therefore$ Molarity

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

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

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

$\left(\because 1 L =1 \,dm ^{3}\right)$
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Concepts Used:

Stoichiometry

Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products, leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of the products can be empirically determined, then the amount of the other reactants can also be calculated.

Stoichiometry helps us determine how much substance is needed or is present. Things that can be measured are;

  1. Reactants and Products mass
  2. Molecular weight
  3. Chemical equations
  4. Formulas

​​Stoichiometric Coefficient

The Stoichiometric coefficient of any given component is the number of molecules and/or formula units that participate in the reaction as written.

Mole Ratios

The mass of one mole of a substance in grams is called molar mass. The molar mass of one mole of a substance is numerically equal to the atomic/molecular formula mass.