2H3PO4 + 3Na2SO4 → 2Na3PO4 + 3H2SO4

A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2

Andrews [41]

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, ΔTf is the depression in freezing point of water.

i is van't Hoff factor.

Kf is the molal depression constant.

m is the molality of the solute.

Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).

van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

4 0

2 years ago

Titanium has five common isotopes: 46Ti (8.0%), 47Ti (7.8%), 48Ti (73.4%), 49Ti (5.5%), 50 Ti (5.3%). What is the average atomic

zhannawk [14.2K]

(46x8.0)+(47x7.8)+(48x73.4)+(49x5.5)+(50x5.3) = 4792.3

4792.3/100 = 47.923 this is the average atomic mass of Titanium

6 0

2 years ago

What is the mass of 0.5 moles of carbon tetrafluoride, CF4?

VashaNatasha [74]

Answer:

44 g

Explanation:

The formula for the number of moles (n) is equal to $n=\frac{mass}{molecular weight}$ .

Since we need to find the mass, we derive it from the formula of the number of moles and we get that mass = n x molecular weight .

The molecular weight of $CF_{4}$ = 12 g/mol (from the carbon) + 19x4 g/mol (from the 4 fluorine atoms)= 88 g/mol

We plug in the numbers in the derived formula for the mass and we get :

mass = n x molecular weight = 0.5 mol x 88 g/mol = 44 g

4 0

2 years ago

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As illustrated, the below manometer consists of a gas vessel and an open-ended U-tube containing a nonvolatile liquid with a den

STALIN [3.7K]

Answer:

1.01atm is the pressure of the gas

Explanation:

The difference in heights in the two sides is because of the difference in pressure of the enclosed gas and the atmospheric pressure. This difference is in mm of the nonvolatile liquid. The difference in mm Hg is:

32.3mm * (0.993g/mL / 13.6g/mL) = 2.36mmHg

As atmospheric pressure is 765mm Hg and assuming the gas has more pressure than the atmospheric pressure (There is no illustration), the pressure of the gas is:

765mm Hg + 2.36mm Hg = 767.36 mmHg

In atm:

767.36 mmHg * (1atm / 760 mmHg) =

1.01atm is the pressure of the gas

5 0

2 years ago

Complete and balance the following redox reaction in acidic solution As2O3(s) + NO3- (aq) → H3AsO4(aq) + N2O3(aq)

scoray [572]

Answer:

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

Explanation:

Oxidation: $As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)$

Balance As: $As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)$
Balance H and O in acidic medium: $As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$
Balnce charge: $As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$ ......(1)

Reduction: $NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$

Balance N: $2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$
Balance H and O in acidic medium: $2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$
Balance charge: $2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$ ......(2)

$Equation (1)+Equation (2)$ gives-

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

3 0

2 years ago

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