Answer:
We know,
\[M=\frac{{{w}_{B}}\times
1000}{{{m}_{B}}\times V}\]
Given : \[{{w}_{B}}\]
= 20 g, \[{{m}_{B}}\] (molar mass of sugar) = 342 g \[mo{{l}^{-1}}\]
\[V=2L=2000mL\]
Putting these values
in Eqn. (i), we get
\[M=\frac{20\times
1000}{342\times 2000}=0.058\]
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