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Question

Chemistry Question on Expressing Concentration of Solutions

Molality of 2.5 g2.5\, g of ethanoic acid (CH3COOH)(CH_{3}COOH) in 75 g75\, g of benzene is

A

0.565 mol kg−10.565\,mol \,kg^{-1}

B

0.656 mol kg−10.656\, mol\, kg^{-1}

C

0.556 mol kg−10.556\,mol\,kg^{-1}

D

0.665 mol kg−10.665\,mol\,kg^{-1}

Answer

0.556 mol kg−10.556\,mol\,kg^{-1}

Explanation

Solution

Molality (m)= mass of solute ( in g)×1000 molecular weight of solute × massofsolvent ( in g)( m )=\frac{\text { mass of solute }(\text { in } g ) \times 1000}{\text { molecular weight of solute } \times \text { massofsolvent }(\text { in } g )} =2.5×100060×7.5=0.555 mol kg−1=\frac{2.5 \times 1000}{60 \times 7.5}=0.555\, mol\, kg ^{-1}