What mass of solute is present in a $0.250 \cdot L$ $0.250L$ volume of $N a O H$ $NaOH$ solution, for which $[N a O H] \equiv 2.50 \cdot m o l \cdot L^{- 1}$ $[NaOH]-=2.50mol*L^-1$ ?

Question

1982 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-30T04:09:26

Approx. $25*g.................$

First we calculate the molar quantity present in the given volume of $NaOH$ ........

$"Molarity"="Moles of solute"/"Volume of solution"...........$

And thus $"Moles of solute"="Volume"xx"molarity"$ .....

$250*xx10^-3*cancelLxx2.5*mol*cancel(L^-1)=0.625*mol$ , with respect to $NaOH$ ...............

which molar quantity corresponds to a MASS of..........

$0.625*molxx40.0*g*mol^-1=??*g$

You must simply KNOW that $1*mL-=1xx10^-3*L$ , i.e. the $"milli"$ prefix $-=10^-3$ .

What mass of solute is present in a 0.250*L0.250⋅L0.250*L volume of NaOHNaOHNaOH solution, for which [NaOH]-=2.50*mol*L^-1[NaOH]≡2.50⋅mol⋅L−1[NaOH]-=2.50*mol*L^-1?