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If 4 g of NaOH dissovles in 36g of H(2)O...

If 4 g of NaOH dissovles in 36g of `H_(2)O`, calculate the mole fraction of each component in the solution. (specific gravity of solution is `1g mL^(-1)`).

Text Solution

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`"4 g NaOH"=(4)/(40)"mole"="0.1 mole"`
`"36 g "H_(2)O=(36)/(18)"mole = 2 mole"`
`therefore " Mole fraction of "H_(2)P=1-0.047=0.953(or(2)/(2+0.1)=0.95)`
`"Total mass of solution "="Mass of solvent "+"Mass of solution"=36+4g=40g`
`"Volume of solution"=("Mass")/("Sp. gravity")=("40 g")/("1 g mL"^(-1))=40mL=0.040L`
`"Molarity of solution "=("Moles of solute")/("Volume of solution in L")=("0.1 mole")/("0.040 L")=2.5M`
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  • When 1.80g glucose dissolved in 90g of H_(2)O ,the mole fraction of glucose is

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