$\begin{array}{1 1}0.617m,0.01,0.99,0.67M\\1.617m,0.02,0.98,0.57M\\2.617m,0.05,0.95,0.47M\\4.617m,0.08,0.92,0.87M\end{array} $

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10g glucose is present in 100g solution (ie) 90g of water =0.090 kg of water

10g glucose =$\large\frac{10}{180}$ mol

$\qquad\qquad\;=0.0555$ mol

90g $H_2O=\large\frac{90}{18}$$=5$ moles

Molarity $=\large\frac{0.0555}{0.090kg}$$=0.617m$

Mole fraction of glucose $x_B=\large\frac{0.0555}{5+0.0555}$

$\Rightarrow 0.01$

Mole fraction of water $(x_A)=1-0.01$

$\Rightarrow 0.99$

100g of solution =$\large\frac{100}{1.2}$mL

$\Rightarrow$ 83.33mL

$\Rightarrow 0.08333$L

Molarity =$\large\frac{0.0555}{0.08333L}$$=0.67$M

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