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3 years ago

How many grams of Na2CO3 must be dissolved into 155g of water to create a solution with a molarity of 8.20 mol/kg

Chemistry

1 answer:

zhannawk [14.2K]3 years ago

4 0

Answer:

135g Na2CO3

Explanation:

I'm going to assume you mean Molality which is mol solute/kg solvent

Molarity would be mol soute/ L solution

we know we have 155g of water which is .155 kg

essentially we have the equation:

mol/kg = 8.20

we substitute .155 in for kg and get:

mol/.155 = 8.20

Solving this gives mol = 1.271

now we must convert to grams using the molar mass

Molar mass Na2CO3 = 106G/mol

so to cancel moles we multiply:

1.271mol x 106g/mol

= 135g

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Read 2 more answers

A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il

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Answer:

The volume of the gas is 2.80 L.

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

P= 2 atm
V= ?
n=0.223 moles
R= 0.0821 $\frac{L*atm}{mol*K}$
T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 $\frac{L*atm}{mol*K}$ * 306 K

Solving:

$V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\$

V= 2.80 L

The volume of the gas is 2.80 L.

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