2.79
58,49 g NaCl---------------------1 N10 g---------------------------------xx= 10/58,49=0,171 N58,49 is the molar mass of NaCl (for the foemula unit).Molality= 0,171/2=0,085
To prepare a 0.9% solution take 0.9grams NaCl and dilute with 100mls of water.
117 grams of NaCl
When you flame test the two solutions, any Na solution burns yellow, while any K solution burns violet/purple. So both the KCl and the KBr will burn purple, while both NaBr and NaCl will burn yellow.
[117(g NaCl) / 58.5(g NaCl/mol NaCl)] / 40.0(L solution) = [117/58.5]/40.0 = 2.00(mol NaCl) / 40.0(L) = 0.0500 mol NaCl / L solution = 0.0500 M
The concentration of NaCl is 263 g/L
The answer is 0,505.
2mol/kg
The density can be measured experimentally.
58,49 g NaCl---------------------1 N10 g---------------------------------xx= 10/58,49=0,171 N58,49 is the molar mass of NaCl (for the foemula unit).Molality= 0,171/2=0,085
58,49 g NaCl---------------------1 N10 g---------------------------------xx= 10/58,49=0,171 N58,49 is the molar mass of NaCl (for the foemula unit).Molality= 0,171/2=0,085
convert the .2 kg of NaCl to moles of NaCl.
Molality(m)= moles of solute divided by kilograms(kg) of solvent you need to find the moles of NaCl by using a conversion factor 70g NaCl = 1mol divided by 58.44g NaCl =1.20mol then you need to convert 300g of water which is the solvent to kilograms by moving the decimal over 3 units to the left which makes .300 kg of solvent Molality= 1.20mol divided by .300kg Molality=4.00
m (molality)=moles of solute/kg of solvent To calculate moles of solute you convert 292.5 g of NaCl to moles of NaCl. So (292.5gNaCl)(1mole NaCl / 58.4gNaCl)= 5.0085 moles NaCl Now plug and chug. 0.25m=5.0085moles NaCl / x x=5.0085 moles NaCl /.25m x=20kg
Convert the 200 mol of water to kilograms of water.
Convert the 200 mol of water to kilograms of water.
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