What mass of sodium chloride is required in the following scenario?

Question

A $350.00 \cdot L$ $350.00L$ volume of $N a C l (a q)$ $NaCl(aq)$ is required at $4.315 \cdot m o l \cdot L^{- 1}$ $4.315mol*L^-1$ concentration. What mass of solute is required?

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Answer 1 · 2017-04-06T06:17:32

You would need over $88 \cdot k g$ $88*kg$ of salt........

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$ .

And thus $moles of solute = concentration \times volume of solution$ $"moles of solute"="concentration"xx"volume of solution"$

And the problem gives us 2 of these 3 quantities, we just have to multiply and divide.

$"moles of solute"=350.00*cancelLxx4.315*mol*cancel(L^-1)=1510.25*mol$ .

And we then need to multiply this molar quantity by the molar mass of $NaCl$ in $g*mol^-1$ :

$1510.25*cancel(mol)xx(35.45+22.99)*g*cancel(mol^-1)=$

$??*g$ of salt..........