# CHE 101 PSU The Percentage Concentration of Any Solution Discussion

• January 31, 2022/

### Description

Question: If 10.0 grams of NaCl is dissolved in 100.0 mL of water what is the concentration in weight/volume percent? What is the molarity?

First:

1. the weight/volume percent is 10%
2. the molarity is 0.855molL-1

1. weight by volume percent is given by the formula (w/v)%=(weight of solute in gram /volume of solution in ml) x 100 given weight of solute =10 gram volume of solution=100ml put these values in above formula (w/v)%=(10g/100ml) x 100% (w/v)%=10%

so the weight/volume percent is 10%

2. molarity is defined as number of moles of solute dissolved per litre of solution represented by the formula molarity =number of moles/volume of solution in litre firstly we have to find out number of moles of NaCl number of moles are given by the formula number of moles =given mass/molar mass given mass of NaCl=5 gram molar mass of NaCl=58.44gmol-1 so number of moles-=5g/58.44gmol-1 number of moles=0.0855mol

so there are 0.0855 mol of NaCl present

given volume of solution =100ml convert this inti litre 1litre=1000ml 1ml=1/1000litre

100ml=100/1000=0.1litre

put these values in the formula of molarity molarity=0.0855mo/0.1L=0.855molL-1

so the molarity is 0.855molL-1

Second:

To calculate the concentration %(mass/volume) and molarity of 10 grams of NaCl that dissolves in 100.0 mL of H2O is done by following these steps:

First find the given information – 10g of NaCl 100.0mL of water

Next find equation(s) that will be needed –

For %(mass/volume): Concentration %(m/v) = Mass(g)/Volume(mL) X 100%

Substitute givensè C%(m/v)= 10g/100.0 mL X 100% = 10% (m/v)

For molarity: M= moles/volume(L)

Looking at the formula, we have volume, but it is in milliliters not liters so we need to convert into liters:

10mL X 1L/1000mL = 0.10 L

Also, the formula has moles and we are given grams, therefore we need to convert grams of NaCl to moles of NaCl. Looking at the atomic masses for Na (22.990) and Cl (35.453), the conversion needed is 58.443g/1 mole. Now we can convert:

10g NaCl x 1 mole NaCl/ 58.443g NaCl = .1701107 è 0.17 mol NaCl

Substitute into the equation:

M = .17 mol NaCl/ 0.10L = 1.7 M

So the answers to this problem of concentration and molarity are 10% concentration(m/v) and 1.7 M for molarity.