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Chemistry Question on Stoichiometry and Stoichiometric Calculations

If the density of methanol is 0.793 kg L–10.793\, \text{kg L}^{–1}, what is its volume needed for making 2.5 L of its 0.25 M solution?

Answer

Molar mass of methanol (CH3OH)(\text{CH}_3\text{OH}) = (1 × 12) + (4 × 1) + (1 × 16)

=32g mol−1= 32 \text{g mol}^{-1}

=0.032kg mol−1= 0.032 \text{kg mol}^{-1}

Molarity of methanol solution =0.793kg L−10.032kg mol−1= \frac{0.793 \text{kg L}^{-1} }{ 0.032 \text{kg mol}^{-1}}

=24.78mol L−1= 24.78 \text{mol L}^{-1}

(Since density is mass per unit volume)

Applying, M1V1=M2V2\text{M}_1\text{V}_1 = \text{M}_2\text{V}_2

(Given solution) (Solution to be prepared)

(24.78 mol L−1)V1=(2.5L)(0.25mol L−1)(24.78\, \text{mol L}^{-1} ) V_1 = (2.5 \text{L}) (0.25 \text{mol L}^{-1} )

V1=0.0252 LV_1 = 0.0252\, \text{L}

V1=25.22 mLV_1 = 25.22\, \text{mL}