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3 years ago

KCl is an ionic salt with a formula weight of 74.54 g/mole.

Chemistry

1 answer:

aalyn [17]3 years ago

5 0

Answer:

mass of water = 958 g = 0.958 kg

m = 1.48 mol/kg

Explanation:

Given:

MM(KCl)= 74.54 g/mol

M= 1.417 mol/L

density of solution= 1064 g/L

To calculate the mass of water in 1 L of solution, first we have to multiply the molarity of the solution (M) by the molecular weight of KCl (MM):

1.417 mol/L x 74.54 g/mol = 105.62 g/L

As the density of the solution is 1064 g/L, we know that 1 L is equal to 1064 g. So, we have 105.62 g of KCl in 1064 g of solution. The mass of water will be the difference between the mass of solution and the mass of solute:

mass of water (g) = 1064 g - 105.6 g = 958.4 g

mass of water (kg)= 958.4 g x 1 kg/1000 g = 0.958 kg

To determine the molality (m):

m = moles of KCl/kg water

m = 1.417 moles/0.958 kg = 1.479 mol/kg

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Which statement compares the masses of two subatomic particles

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3 years ago

Phenol, c6h5oh, is a stronger acid than methanol, ch3oh, even though both contain an o]h bond. Draw the structures of the anions

timurjin [86]

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3 years ago

Intravenous, or IV, solutions used in medicine must exert the same osmotic pressure as blood to prevent a net flow of water into

Novosadov [1.4K]

Answer:

$\boxed{\text{7.5 bar}}$

Explanation:

1. Find the molar concentration

Assume that we have 100 mL solution.

Mass of NaCl = 0.90 g

$\text{Moles of NaCl} = \text{0.90 g} \times \dfrac{\text{1 mol}}{\text{58.44 g}} = 0.0154 \text{ mol}\\\\c = \dfrac{\text{Moles of NaCl}}{\text{Litres of solution}} = \dfrac{0.0154 \text{ mol}}{\text{0.100 L}} = \text{0.154 mol/L}$

2. Find the osmotic pressure

The formula for osmotic pressure (Π) is

Π = icRT

T = (37 + 273.15) K = 310.15 K

$\Pi = 1.88 \times \text{0.154 mol}\cdot \text{L}^{-1} \times \text{0.083 14 bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K} = \text{7.5 bar}\\\\\text{The osmotic pressure of blood at body temperature is }\boxed{\textbf{7.5 bar}}$

3 0

3 years ago

A pair of students found the temperature of 100. g of water to be 26.5°C. They then dissolved 8.89 g of AgNO3 in the water. When

bekas [8.4K]

Explanation:

Initial temperature of the water = $T_i=26.5^oC$

Final temperature of the water = $T_f=23.7^oC$

a) Change in temperature of the water,ΔT = $T_f-T_i$

$\Delta T=23.7^oC-26.5^oC=-2.8^oC$

Change in temperature of the water,ΔT is -2.8°°C.

b) Endothermic reaction : Reaction in which heat absorbed and the temperature if the surrounding is decreased.

Exothermic reaction : Reaction in which heat released and the temperature if the surrounding is increased.

On dissolving silver nitrate in water the temperature of the water decreased 2.8 degree Celsius which means that energy water was absorbed by solid silver nitrate to get dissolve in water. Hence, endothermic reaction.

(f) Mass of silver nitrate = 8.89 g

Moles of silver nitrate = $\frac{8.89 g}{170 g/mol}=0.05229 mol$