One liter of a one molar solution of NaOH in water contains 40g of NaOH. The quantity must be known.
The solution has a molar concentration of 0,33.
Try 1 molar sodium hydroxide NaOH solution and 1 molar hydrochloric acid solution HCl . Mix them together and you have a common salt solution NaCl in water. Not exactly sure what you wanted, but there you go.
10.0 ml of a solution of 6.0 molar NaOH contains (10.0/1000)(6.0) = 0.060 moles of NaOH. Call the number of milliliters needed to make a solution that is 0.30 molar with this amount of NaOH v. Then 0.060/v = 0.30, or v = 50 ml. Ideally, this solution should be made by putting the 6.0 molar solution into a volumetric container calibrated to 50 ml, then diluting with water to the mark for 50 ml. As an approximation, 50 - 10.0 = 40 ml of water can be added, but this may not give an exact match because 6.0 molar NaOH solution probably has a density considerably higher than 1.00 grams/ml, the density of water at standard temperature and pressure, while the density of 0.30 molar solution is substantially less but may still not be as low as 1.00. Therefore, adding a specified volume of water may not result in a total volume equal to the sum of the volume of water added plus the 10.0 ml originally specified.
There is 0.5 moles of NaOH per litre To calculate 0.5 molar NaOH first know the molecular weight of NaOH i.e 40 now multiply the number of moles of NaOH you have (0.5) found as above. so to find the number of grams of NaOH we needed to start with (0.5) * (40) = 20g So dissolve 20g of NaOH in one litre of the solution to prepare 0.5 molar solution
Dissolve 40 grams of solid NaOH into enough water that it dissolves completely. Transfer this to a volumetric flask. Then add water until the volume of the solution is 1 liter. Care is needed as heat is generated which affects the volume. It is best to make up with water to about 2 cm short of the mark, allow to cool and then add the balance of the water.
The solution has a molar concentration of 0,33.
6 molar
Try 1 molar sodium hydroxide NaOH solution and 1 molar hydrochloric acid solution HCl . Mix them together and you have a common salt solution NaCl in water. Not exactly sure what you wanted, but there you go.
It is 2.5 molar. The reason for this is that molarity means moles per litre. You have to multiply by 5 to get from 200ml to a litre, so you have to do the same with the moles.
10.0 ml of a solution of 6.0 molar NaOH contains (10.0/1000)(6.0) = 0.060 moles of NaOH. Call the number of milliliters needed to make a solution that is 0.30 molar with this amount of NaOH v. Then 0.060/v = 0.30, or v = 50 ml. Ideally, this solution should be made by putting the 6.0 molar solution into a volumetric container calibrated to 50 ml, then diluting with water to the mark for 50 ml. As an approximation, 50 - 10.0 = 40 ml of water can be added, but this may not give an exact match because 6.0 molar NaOH solution probably has a density considerably higher than 1.00 grams/ml, the density of water at standard temperature and pressure, while the density of 0.30 molar solution is substantially less but may still not be as low as 1.00. Therefore, adding a specified volume of water may not result in a total volume equal to the sum of the volume of water added plus the 10.0 ml originally specified.
The answer is 0,625 moles.
Molarity is moles per litre. So here, you have to divide by 353ml, x 1000 (to get it per litre). .73/353*100 is 2.07 molar.
There is 0.5 moles of NaOH per litre To calculate 0.5 molar NaOH first know the molecular weight of NaOH i.e 40 now multiply the number of moles of NaOH you have (0.5) found as above. so to find the number of grams of NaOH we needed to start with (0.5) * (40) = 20g So dissolve 20g of NaOH in one litre of the solution to prepare 0.5 molar solution
Dissolve 40 grams of solid NaOH into enough water that it dissolves completely. Transfer this to a volumetric flask. Then add water until the volume of the solution is 1 liter. Care is needed as heat is generated which affects the volume. It is best to make up with water to about 2 cm short of the mark, allow to cool and then add the balance of the water.
10kg of water contains 10l of water. So morality is 0.2mildm-3.
The molar mass remain the same but because the NaOH solution absorb carbon dioxide the titer of this standardized solution was altered.
It is possible only if you evaporate the water.