Mole fraction of glucose in aqueous solution = 0.1

Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9.

So mass % of water = ${{0.9 \times 18} \over {0.1 \times 180 + 0.9 \times 18}}$ $\times$ 100

= 47.37 $\simeq$ 47