Question

A patient suffering from deficiency of sodium, was given intravenous 1 L of 0.9%(w/v) sodium chloride (normal saline) solution. The amount of sodium chloride needed to prepare this 1 lit of solution is ‘x’ g. Find ‘x’. Sarika dissolved this ‘x’ g of sodium chloride in 1 L of water to prepare the normal saline solution in the school laboratory, but her teacher said that she has made a mistake. Can you identify what mistake has been done by Sarika?

please give a proper answer !!​

Answer 1

Answer:

The concentration of the sodium chloride (NaCl) solution is given as 0.9% (w/v), which means that 0.9 g of NaCl is present in 100 mL of the solution. To prepare 1 L (1000 mL) of the solution, we need to use the proportion:

0.9 g / 100 mL = x g / 1000 mL

Solving for x, we get:

x = 8.1 g

Therefore, 8.1 g of NaCl is needed to prepare 1 L of 0.9% (w/v) sodium chloride solution.

The mistake made by Sarika is that she dissolved the 8.1 g of NaCl in 1 L of water, which would result in a concentration of only 0.81% (w/v) sodium chloride solution. To prepare a 0.9% (w/v) solution, she should have dissolved the 8.1 g of NaCl in 900 mL of water and then made up the volume to 1 L.