# What mass of "nickel(II) chloride hexahydrate" is present in a 250*mL volume of 1.00*mol*L^-1 concentration?

Jan 8, 2018

By definition, $\text{molarity"="moles of solute"/"volume of solution}$...we require a mass of approx. $75 \cdot g$

#### Explanation:

Here you specify a $\text{molarity} = 1.00 \cdot m o l \cdot {L}^{-} 1$ with respect to $N i {\left(N {O}_{3}\right)}_{2} \cdot 6 {H}_{2} O$ and a volume of $250 \cdot m L \equiv 0.250 \cdot L$...

And so we require a molar quantity of....

$\text{moles of solute"="molarity"xx"volume}$

And note that the given product has units of $m o l \cdot {L}^{-} 1 \times L \equiv \text{moles}$ as required....

And so .........

$\text{moles of nickel salt hydrate} = 1.00 \cdot m o l \cdot {L}^{-} 1 \times 0.250 \cdot L = 0.250 \cdot m o l$

And to get the mass we take the product....

0.250*molxxunderbrace(290.79*g*mol^-1)_"molar mass of metal hydrate"=??*g