First calculate the mass you need from NaCl I will calculate it for you now: 100 cm3= 0,100 L n (NaCl) = 0,1 mol/dm3 * 0,100 L = 0,01 mol m (NaCl) = n*Mm = 0,01 mol * ( 22,99+35,45) g/mol = 0,01 mol * 58,44 g/mol = 0,5844 g You will need 0,5844 g NaCl and you will mix it woth 100 cm3 water
To prepare a 2% NaCl (w/v) solution, you would dissolve 2 grams of NaCl in enough water to make 100 mL of solution. This means you would add 2 grams of NaCl to a flask and then add water until the total volume reaches 100 mL.
To determine which solution is more concentrated, we need to calculate the concentration of each solution in grams per cubic centimeter (g/cm3). For the first solution with 20g of alkali in 250 cm3 of solution, the concentration would be 20g / 250 cm3 = 0.08 g/cm3. For the second solution with 10g of alkali in 500 cm3 of solution, the concentration would be 10g / 500 cm3 = 0.02 g/cm3. Therefore, the first solution with 20g of alkali in 250 cm3 of solution is more concentrated at 0.08 g/cm3 compared to the second solution with 10g of alkali in 500 cm3 of solution at 0.02 g/cm3.
right i dont know this 4 sure but because u want a 0.1 mol/dm3 and u only need 100cm3 u will need 0.01mols of copper sulfate to dilute in 100cm3. soo now u have a solution that is 0.01mols per 100cm3 or 0.1 mols per 1000cm3 (dm3)
The number of grams is the number of cm3. e.g. 300g = 300cm3
Because its molecules (sodium bound ionically to chlorine) can spread out within the water (H2O) molecules. The polar water molecules attract both the sodium (positive ions) and the chlorine (negative ions). You may be interested to know that you can dissolve 34 grams of salt in 100 cm3 of water at 10 degrees C.
To prepare a 2% NaCl (w/v) solution, you would dissolve 2 grams of NaCl in enough water to make 100 mL of solution. This means you would add 2 grams of NaCl to a flask and then add water until the total volume reaches 100 mL.
This density is 1,0707 g/cm3 at 20 0C.
- freezing point for a solution of 35 g/L NaCl: -2 deg. Celsius- density for a solution of 35 g/L NaCl: 1,025 g/cm3- thermal conductivity for a solution of 35 g/L NaCl: 0,6 W/m.K
- freezing point for a solution of 35 g/L NaCl: -2 deg. Celsius- density for a solution of 35 g/L NaCl: 1,025 g/cm3- thermal conductivity for a solution of 35 g/L NaCl: 0,6 W/m.K
The answer is 0,1648 g NaCl.
Dissolve 0.1g of the powder in ethanol and dilute with distilled water upto 100 cm3
To determine which solution is more concentrated, we need to calculate the concentration of each solution in grams per cubic centimeter (g/cm3). For the first solution with 20g of alkali in 250 cm3 of solution, the concentration would be 20g / 250 cm3 = 0.08 g/cm3. For the second solution with 10g of alkali in 500 cm3 of solution, the concentration would be 10g / 500 cm3 = 0.02 g/cm3. Therefore, the first solution with 20g of alkali in 250 cm3 of solution is more concentrated at 0.08 g/cm3 compared to the second solution with 10g of alkali in 500 cm3 of solution at 0.02 g/cm3.
The density of sodium chloride is 2,165 g/cm3.
The sodium chloride density is 2,165 g/cm3 and the melting point is 801 0C.
1 m = 100 cm So 1 m3 = 100*100*100 cm3 = 106 cm3 and 10 m3 = 107 cm3
volume given =27.5mlmolarity of base=0.105find number of moles=moles=vol(cm3)/1000*molarity of NaClmoles=27.5/1000*0.105=3.82 moles of sodium chloride
The concentration of a solution is moles/volume. 2.943g of pure sulphuric acid H2S4 is approximately 15 moles, and 15/150 cm3 is 10.