What is the mass of the solute in 250 mL of 0.0520 M $N a O H$ $NaOH$ solution?

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Answer 1 · 2016-06-16T06:28:09

$Moles of solute = Volume \times Concentration$ $"Moles of solute"="Volume"xx"Concentration"$ ;

$Mass of solute = Moles \times Molar mass .$ $"Mass of solute"="Moles"xx"Molar mass".$

This relationship is clear dimensionally.

So $moles of solute =$ $"moles of solute "=$

$250 \cdot m L \times 10^{- 3} L \cdot m L^{- 1} \times 0.0520 \cdot m o l \cdot L^{- 1}$ $250*mLxx10^-3L*mL^-1xx0.0520*mol*L^-1$

$= 0.0130 \cdot m o l$ $=0.0130*mol$

We have $0.0130 \cdot m o l$ $0.0130*mol$ sodium hydroxide.

The mass of caustic soda is simply this quantity mulitiplied by the molar mass of caustic:

$0.0130 \cdot m o l \times 40.00 \cdot g \cdot m o l^{- 1}$ $0.0130*molxx40.00*g*mol^-1$ $=$ $=$ $? ? g$ $??g$

What is the mass of the solute in 250 mL of 0.0520 M NaOHNaOH solution?