Chemistry, 15.04.2020 05:14 kelly3215

An unknown diprotic acid requires 28.20 mL of 0.104 M NaOH to completely neutralize a 1.400-g sample. Calculate the approximate molar mass of the acid.
A. 239 g/mol
B. 341 g/mol
C. 682 g/mol
D.955 g/mol
E. 477 g/mol

Answers: 2

Show answers

Answers

Answer from: Quest

football

explanation:

because it is

Answer from: Quest

2licl + 3o2 i yook the test and it was correct.

Answer from: Quest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

Answer from: Quest

Answershower asap; wash for 15 min. take off all clothing.

Another question on Chemistry

Chemistry, 21.06.2019 16:00

Review the branily terms and services guides well u know what i never did so go have a nice ice cream sunday

Answers: 1

Answer

Chemistry, 22.06.2019 09:30

In 2002, the rare earth elements mine in mountain pass, california was closed because

Answers: 1

Answer

Chemistry, 22.06.2019 16:00

How could a student test the effect of removing heat from a gas that is stored in a sealed container? what must occur in order for matter to change states?

Answers: 2

Answer

Chemistry, 22.06.2019 22:30

Amedication is given at a dosage of 3.000 mg of medication per kg of body weight. if 0.1500 g of medication is given, then what was the patient's weight in pounds (lbs)? there are 453.59g in 1 lb.

Answers: 2

Answer

You know the right answer?

An unknown diprotic acid requires 28.20 mL of 0.104 M NaOH to completely neutralize a 1.400-g sample...

Questions