Remember the moles equations.
moles = mass(g) / Mr = [conc] x vol(mL) / 1000
Hence
mass(g) = [conc] x vol(mL) x Mr / 1000 (be algebraic rearragnement)
conc = 0.05 M
Mr = 23 + 35.5 = 58.5
vol(mL) = mass (g) = 150 g =150 mL (NB THis is the standard for water at STP)
Substituting
mas9g) = 0.05 x 58.5 x 150 / 1000
mass(g) = 0.43875 g ~ 0.44 g
The term molecule is not adequate for an ionic compound; correct is formula unit.
150 g NaCl contain the equivalent of 2,5667 formula units.
To prepare a 1,5 M sodium chloride solution we need 87,659 g NaCl.
The answer is 87,66 g NaCl.
The answer is 0,44 g NaCl.
This mass is 0,438 g.
It depends on the final solution Volume you want to prepare. For 100ml of a 6M NaCL solution, you add 35.1g of NaCl to water until you reach 100ml. Dissolve and autoclave for 15 mins.
Dissolve 100 mg NaCl in 1.0 Litre water.
Dissolve 2g of NaCl in 100 cm3 water at normal temperature.
Dissolve 30g of sodium chloride in 100 mL of water.
Dissolve 39,665 g NaCl in 1 kg water.
Dissolve 12 g dried sodium chloride (reagent grade) in 100 mL demineralized water.
No, it is not possible.
0.02/5=0.004
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.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
You have to evaporate (by open boiling) 45 mL of the 75 mL 2M NaCl solution thus reducing the volume to 30 mL 5M NaCl.
Molar mass of NaCl =~58.4 g/mole0.1 N NaCl = 0.1 moles/liter To make 1 liter of 0.1N NaCl thus requires 0.1 moles/liter = 0.1 moles x 58.4 g/mole = 5.84 moles Dissolve 5.84 g (6 g using 1 sig. fig.) in a final volume of 1 liter to make 0.1N NaCl