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1 year ago

What mass of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

Chemistry

1 answer:

iren [92.7K]1 year ago

3 0

91.4 grams

91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

C = mol/volume

2.45M=mol/0.5L

2.45M⋅0.5L = mol

mol = 1.225

Convert no. of moles to grams using the atomic mass of K + Cl

1.225mol * $\frac{39.1+35.5}{mol}$

mol=1.225

=1.225 mol . $\frac{74.6g}{mol}$

=1.225 . 74.6

=91.4g

therefore, 91.4 grams of kcl in grams must be added to 500 ml of a 0. 15 m kcl solution to produce a 0. 40 m solution.

What is 1 molar solution?

In order to create a 1 molar (M) solution, 1.0 Gram Molecular Weight of the chemical must be dissolved in 1 liter of water.

58.44 g make up a 1M solution of NaCl.

To learn more about molar solution visit:

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3 years ago

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

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1. the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c ( $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$ )

Explanation:

To find -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa

⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa

⇒Absolute pressure (p2) = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. $V_{1}$ = $V_{2}$

As we know that, P ∝ M

⇒ $\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378$ ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒ $\frac{P}{T}$ ∝ M

⇒ $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$

So, The correct relation is c option.

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