Percentage solution

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In biology, percentage solutions are often preferred to molar solutions. A 1% solution would have 1 g of solute dissolved in a final volume of 100 mL of solution. This would be labeled as a weight/volume [w/v] percentage solution. For w/w, both solvent and solute would need to be weighed in the required ratios. Volume would accordingly be measured using a measuring cylinder, volumetric flask, pipette or similar. Labels should show what the percentage relationships are (w/v, w/w or v/v).

This is done by dividing the weight of the product in g with the volume in mL. Therefore the equation is (Mass(g) / Volume(mL))x100 = %

The molarity of a percentage solution (w/v) can be calculated using the molar mass of the solute used. For example, sucrose (table sugar) has a molar mass of a 342.34 g/mol. A 1% sucrose solution (w/v), therefore, is 1 g / 0.1 L / 342.34 g/mol = 0.029 mol/L, or 29 mM. The concentration in gram per litre is easy to calculate: a 1% solution contains 1 g per 100 mL or 10 g per 1000 mL (1 L); therefore, the percentage is to be multiplied by 10.
It is common practice in lab. to make solution with specific concentration from solvent and solute. The following section is related to making percentage solution with unit mass/100ml.

Mass percentage g/100ml calculation

There are three cases in practical calculations to determine right amount of solvent/solute for specific concentration. They are

• Case 1: amount of solvent volume is given.
• Case 2: amount of solute mass is given.
• Case 3: amount of final solution volume is given.

In the following equations, A is solvent, B is solute, and C is concentration. Solute volume contribution is considered through ideal solution model.

• Case 1: amount (ml) of solvent volume V_A is given. Solute mass m_B = C V_A d_A /(100-C/d_B)
• Case 2: amount of solute mass m_B is given. Solvent volume V_A = m_B (100/C-1/ d_B )
• Case 3: amount (ml) of final solution volume Vt is given. Solute mass m_B = C Vt /100; Solvent volume V_A=(100/C-1/ d_B) m_B

All these cases have been covered in solution calculator to help users get results quickly and accurately.
Example: Make 2 g/100ml of NaCl solution with 1 L water with Water_(properties) d=0.997048g/ml, Sodium_chloride density=2.165g/ml.
This is case 1, NaCl mass required is m_B = C V_A d_A /(100-C/d_B)
m_B = 2 x 1000 x 0.997048/ (100 – 2/2.165) =20.1269 gram
If ignore solute effect to solution volume which is applicable at low concentration, the calculation can be simplified as

• Case 1: solvent volume is known, m_B = C V_A /100
• Case 2: solute mass is known, V_A = m_B 100/C
• Case 3: total solution volume is know, same equation as case 1. V_A=Vt; m_B = C V_A /100

Example: Make 2 g/100ml of NaCl solution with 1 L water Water_(properties). Now density information is not required.
m_B = C V_A /100 = 2 x 1000 /100 =20 gram

References:

• A textbook of science for health professional, By Barry Hinwood, 2nd Edition,1992, publisher:Nelson Thurnes Limited, ISBN-13:9780748733774
• A to Z of Thermodynamics Pierre Perrot ISBN 0-19-856556-9

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