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?
Answers
Answer:
Answer:
2y + ( \frac{6}{4} - \frac{3}{5} ) = 5y - \frac{4}{8}2y+(
4
6
−
5
3
)=5y−
8
4
2y + \frac{9}{10} = 5y - \frac{4}{8}2y+
10
9
=5y−
8
4
7y = \frac{14}{10}7y=
10
14
y = \frac{1}{5}y=
5
1
hope it helps
please Mark as brainliest ❤️
- value of x is 9 grams.
- sarika should have made 1 l liter of normal saline with 700 ml of distilled water and 300 ml of water
Given:
Total volume of solution: 1 L
Amount of NaCl = 0.9% (w/v)
TO FIND:
Amount of sodium chloride needed to make 1 L of normal saline solution.
SOLUTION:
We can simply solve the problem as under
100 ml of solution contains 0.9 grams of NaCl
In, 1 mL of the solution has = 0.9/100 grams of NaCl
so, 1000mL (1litre) of solution contains,
= (0.9/100)×1000
Hence, X = 9 grams.
Hence, the amount of sodium chloride needed to prepare 1 lit of NaCl solution is 9 grams.
In the second part of the problem, Sarika wants to make 1 liter of normal saline solution.
To make 1 liter of normal saline-
- Dissolve 9 grams of NaCl in 700ml distilled water.
- Add water to bring the solution to 1000ml.
Hence, Sarika should have added 9 grams of NaCl in 700 ml of distilled water and then add 300 ml of water to bring the solution to 1-liter
#spj3