Question

For the preparation laboratory of a 4% solution (weight / volume) sodium hydroxide at from 50 mL of a 5 mol solution. L-1 this basis, it is necessary(explain your answer) Data: molar mass (g mol-1): Na = 23 O = 16 H= 1

a) adding 200 ml of water. b) adding 250 ml of water. c) adding 500 ml of water. d) evaporating 200 mL of water. e) evaporating 250 mL of water.

Answer 1

First, is important that you know equations involved:

%w/v (w:weight: v:volume)

(EQ.1)

For preparation of a solution for dilution (as in our case), you will have the same moles or grams of solute but with a greater amount of solvent after dilution. Follow equation represents mentioned:

g of solute before dilution= g of solute after dilution ( before=condition 1; after=condition 2)

g1=g2

If use EQ 1

%w/v (1) x v solution (1) = %w/v x v solution (2) (EQ. 2)

CONDITIONS BEFORE DILUTION

V=50mL

M=5M

Calculating g de NaOH in the solution, we must use the concentration and molecular mass:

Masa Molar for NAOH= Atomic mass Na + Atomic Mass O + Atomic Mass H= (23+16+1)g/mol

Molar Mass for NAOH= 40g/mol

M=mol/L

For calculate g of NaOH use Molar Mass

If solution 1 have 10g of NaOH and 50mL in volume, it %w/v is (Applying Eq 1):

SOLUTION 1, IS 20%M/V OF NaOH WITH A VOLUME OF 50mL

CONDITIONS AFTER DILUTION

APPLYING EQ 2, we will know the volume of the resulting solution to achieve a concentration of 4% w / v

%w/v (1) x mL solution (1) = %w/v (2) x mL solution (2)

v(2)=(20% x 50mL)/4%=250mL

FINAL SOLUTION WILL DO OF 250mL. FOR PREPARED SOLUTION 4%w/v OF NaOH FROM A SOLUTION OF 50mL 5M, WE SHOULD ADD 200mL FOR OBTAIN VOLUME CALCULATED (250mL)

ANSWER: adding 200mL of water (OPTION A)

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