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How much of NaCl is in 1.84 L of 0.200 M NaCl Answer in units of mol.?
To find the amount of NaCl in 1.84 L of 0.200 M NaCl, you use the formula: amount = concentration x volume. Thus, amount = 0.200 mol/L x 1.84 L = 0.368 mol of NaCl.
What is 8.44 mci l-131?
8.44 mci l-131 = -122.56
How much salt is saltwater?
The sea water contain approx. 35 g/L NaCl.
What is the molarity of a solution made by dissolving 58.44 g of NaCl in enough water to make 1.0 L of solution?
Divide grams (mass) by molar mass to find moles58.44 (g NaCl/L) / [22.99+35.45](g NaCl/mol NaCl)= 1.000 mol/L NaCl
How much nacl is required for a 200 mM NaCl solution?
You need 58,44 mg of ultrapure NaCl; dissolve in demineralized water, at 20 0C, in a thermostat, using a class A volumetric flask of 1 L.
Why is 0.9 NaCl osmolarity 300 mOsmL calculations?
- 0,9 g/L NaCl equal to 0,154 moles - but because NaCl is dissociated in two ions in water the relation is 1 mol NaCl equal 2 osmol/L - and so 0,9 % NaCl equal 308 milliosmole/L
What is the concentration of 58.5g of NaCl in 2 L of solution?
The concentration of the NaCl solution is 29.25 g/L. This is calculated by dividing the mass of NaCl (58.5 g) by the volume of the solution (2 L).
What is the concentrate of 2M NaCl?
2M NaCl is a equivalent to a solution with 116,88 g NaCl in 1 L water.
How to calculate the percentage of NaCl to M and mOsm?
9 g/L NaCl = 0,9 g/100 mL = 0,154 moles/L = 300 mOsm/L
1.75mol of NaCl in 15.2L of solution?
This is equivalent to a concentration of 6,73 g/L NaCl.
A solution of nacl and water has a molality of 4.5m?
The concentration of NaCl is 263 g/L
How to prepare a 50liters of sea water solution of NaCl how much NaCl is required in gram?
The concentration of NaCl in sea water is about 35 g per kg. The density of seawater is about 1.025 kg/l, so you would need 50 * 1.025 kg sea water for 50 liters of solution.NaCl contributes to this with 35 * 1.025 * 50 g / kg * kg / l * l. consequently the result is about 1794 g of NaCl.
