If 10% (w/w) aqueous solution of K2CO3 is kept in a container whose volume capacity is 100 L then, mole fraction (xK2CO3) is equal to:
A
0.0188
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B
0.0368
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C
0.0254
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D
0.0138
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Solution
The correct option is D 0.0138 10 %(w/w) means 10 g of K2CO3 is present in 100 g of solution. Mass of K2CO3 in 100 g solution = 10 g Number of moles of K2CO3 in 100 g solution =givenmassmolarmass=10138=0.07mol Weight of solvent (water) in 100 g solution =100−10=90g Number of moles of water in 100 g solution =9018=5mol Molefraction=molesofK2CO3molesofK2CO3+molesofsolvent MolefractionofK2CO3=0.07(0.07+5)=0.0138