When 1.80 g of glucose is dissolved in 90 g | vedclass

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Question

(B) Step $1$: Calculate the number of moles of glucose $(n)$. $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{1.80 \ g}{180 \ g/mol} = 0.01 \ mol$.

Vedclass · Accepted Answer

$0.00199$

Vedclass · Answer

(B) Step $1$: Calculate the number of moles of glucose $(n)$. $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{1.80 \ g}{180 \ g/mol} = 0.01 \ mol$. Step $2$: Calculate the number of moles of water $(N)$. $N = \frac{	ext{mass}}{	ext{molar mass}} = \frac{90 \ g}{18 \ g/mol} = 5 \ mol$. Step $3$: Calculate the mole fraction of glucose $(x_{glucose})$. $x_{glucose} = \frac{n}{n + N} = \frac{0.01}{0.01 + 5} = \frac{0.01}{5.01} \approx 0.00199$.

When $1.80 \ g$ of glucose is dissolved in $90 \ g$ of $H_2O$,the mole fraction of glucose is:

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