# How do you calculate the number of grams of solute in 250.0 mL of 0.179 M KOH?

Jul 14, 2017

We use the formula.....$\text{Concentration"="Moles of solute"/"Volume of solution}$, and get approx. $2.5 \cdot g$ $K O H$.

#### Explanation:

And so (i) we calculate the molar quantity.....

$\text{Moles of KOH"="concentration"xx"volume} = 0.179 \cdot m o l \cdot {L}^{-} 1 \times 250.0 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 = 4.475 \times {10}^{-} 2 \cdot m o l .$

And then (ii) we multiply this molar quantity by the formula mass of $K O H$, .....4.475xx10^-2*molxx56.11*g*mol^-1=??*g

Can you tell us $p O H$ and $p H$ of this solution...........?