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The stock concentration of NaOH is 0.5M. 0.5M = 0.500mol/1.0L.
a) Calculate the mole of NaOH required to prepare 200mL of 0.5M solution.
b) How many grams of NaOH required from the number of mole of NaOH that you calculate from9a. The MW of NaOH = 40.0g/mol
c)Calculate % (m/v) of NaOH from the above. m/v means mass of solute/100mL of solution. Hint: The amount of gram calculated above is based on 200 mL, how many grams should be if you only have 100mL
d) Assume the density of NaOH is 1.00g/mL How much water is required to be added to prepare the above solution? Remember your total volume is 200 mL.

Answer :

IHLESTERGO

Answer:a) 0.1 mole. b) 4g. c) 2% d)  196 mL

Explanation: in 200mL , 0.1mole

mw NaOH = 40g/mol —> 4g in 0.1 mole

4g in 200mL so 2g in 100mL

density NaOH = 1g/mL so if 4g in 200 mL,  4mL , 196 mL water

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