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Chemistry Question on Solutions

If the density of methanol is 0.8 kg L−10.8\, kg\, L ^{-1}, what is its volume needed for making 4 L4\, L of its 0.25 M0.25\, M solution?

A

4 mL

B

8 mL

C

40 mL

D

80 mL

Answer

40 mL

Explanation

Solution

Given, density of CH3OH=0.8 kg L−1CH _{3} OH =0.8\, kg\, L ^{-1}
Molarity =0.25 M=0.25\, M. Volume of 0.25 M=4L0.25\, M =4 L
Volume needed =?=?
First of all we find mass of methanol (i.e. given mass)
Molarity = Number of moles  Volumeof solution(L) =\frac{\text { Number of moles }}{\text { Volumeof solution(L) }}
Molarity = Given mass  Molar mass ×1 Volume of solution(L) =\frac{\text { Given mass }}{\text { Molar mass }} \times \frac{1}{\text { Volume of solution(L) }}
Molar mass of CH3OH=12+3+16+1=32 mol−1CH _{3} OH =12+3+16+1=32\, mol ^{-1}
∴0.25 mol−1= Given mass 32 g mol−1×14L\therefore 0.25\, mol ^{-1}=\frac{\text { Given mass }}{32\, g\,mol ^{-1}} \times \frac{1}{4 L}
∴\therefore Given mass =32 g=32\, g or 0.032 kg0.032\, kg
Again ∵\because denstiy
= given mass (kg) volume (mL)=\frac{\text { given mass }( kg )}{\text { volume }( mL )}
∴0.8 kg L−1=0.032 kgV(mL)\therefore 0.8\, kg\, L^{-1}=\frac{0.032\, kg }{V( mL )}
or 0.8×1000kg mL−1=0.032 kgV(mL)0.8 \times 1000 kg\, mL ^{-1}=\frac{0.032\, kg }{V( mL )}
∴V(mL)=0.032 kg0.8 kgL−1×1000 V(mL)=40\therefore V(m L)=\frac{0.032\, kg }{0.8\, kg L^{-1}} \times 1000\, V (m L)=40