Question

1) Sodium hydroxide is deliquescent. A sample of 3.0 g was dissolved in 100 mL; 10 mL was titrated with 34.9 mL HCl 0.2 M. What is the percent of absorbed water in the sample? ​

Answer 1

From the calculation, the percentage of water in the sodium hydroxide sample is 7%.

What is neutralization?

The term neutralization has to do with the reaction between an acid and a base to yiled salt and water.

Now we have  to apply the titration formula;

CAVA/CBVB = NA/NB

CA = concentration of acid

CB = concentration of base

VA = volume of acid

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

The reaction equation is; HCl + NaOH ----->NaCl + H2O

CAVANB = CBVBNA

CB = CAVANB /VBNA

CB = 34.9 * 0.2 M * 1/10 * 1

CB = 0.698 M

Number of moles = Conncentration * volume

= 0.698 M * 100/1000 L = 0.0698 moles

Mass = Number of moles * molar mass

Mass =  0.0698 moles * 40 g/mol = 2.79 g

percent of NaOH = 2.79 g/ 3g * 100/1 = 93%

Percent of water = 100- 93 = 7%

Learn more about neutralization: brainly.com/question/15395418