A $16 \cdot g$ $16g$ mass of sodium hydroxide is dissolved in enough water to give $2.0 \cdot L$ $2.0L$ of solution. What is the concentration of the solution in $m o l \cdot L^{- 1}$ $mol*L^-1$ ?

Question

1973 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-02T19:40:52

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

$Molarity$ $"Molarity"$ $=$ $=$ $((16.0*cancelg)/(40.0*cancelg*mol^-1))/(2.0*L)=0.20*mol*L^-1$ .

Note how the units cancel to give the required units of concentration, i.e. $mol*L^-1$ :

$((cancelg)/(*cancelg*mol^-1))/(L)=mol*L^-1$ , i.e. $1/x^-1$ $=$ $1/(1/x)$ $=$ $x$ as required...........

A 16*g16⋅g16*g mass of sodium hydroxide is dissolved in enough water to give 2.0*L2.0⋅L2.0*L of solution. What is the concentration of the solution in mol*L^-1mol⋅L−1mol*L^-1?