What volume of $"lithium sulfite solution"$ of $2.50molL^-1$ concentration would contain a mass of $422.55*g$ ?

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Answer 1 · 2017-04-17T20:14:47

$"Moles of lithium sulfite req'd..........."=(422.55*g)/(93.94*g*mol^-1)=$

$=(422.55*g)/(93.94*g*mol^-1)=4.50*mol$ .

And since $"concentration"="moles"/"volume"$ , we need the quotient..........

$"volume"="moles"/"concentration"=(4.50*mol)/(2.50*mol*L^-1)=1.80*1/L^-1$

$1.80*1/(1/L)=1.80*L$

In practice, water is quite an involatile material. You would not evaporate this quantity of water.

What volume of "lithium sulfite solution""lithium sulfite solution" of 2.50*mol*L^-12.50*mol*L^-1 concentration would contain a mass of 422.55*g422.55*g?