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Question

A student prepares a solution by dissolving 3.72 g of NaOH in enough water to make 300. mL of solution.
The student then transfers 32.0 mL of this solution into a 100. mL volumetric flask. Water is then added
up to the 100. mL mark.
What is the final molarity of NaOH in the 100. mL flask ?

Answer 1

Atomic weights: Na=23 O=16 H=1 NaOH=40

Let the first NaOH solution be called X. Let the second NaOH solution be called Y.

3.72gNaOH/300mLX x 1molNaOH/40gNaOH x 1000mLX/1LX = 0.31 mole/L

32mLX/100mLY x 0.31molNaOH/1000mLX x 1000mLY/1LY = 0.099 mole/L

Answer 2

300 ml are equivalent to------- 3.72 g

32 ml are equivalent to-------- 3.72(32)/300 = 0.4g

since 32 ml were made upto 100 ml, now 100 ml contains 0.4 g

so 1000 ml contains (0.4)*10 = 4 g

Molecular weight of NaOH = 40 g

40 g in 1000 ml = 1 molarity

so (40/10=) 4 g in 1000 ml = 1/10 = 0.1 molarity

so final molarity of NaOH in 100 ml flask = 0.1

Answer 3

you have to cross the mole bridge-lol