Use M1V1=M2V2
Where M is the concentration (5.5 m for M1 and 1.2 m for M2) and V is volume V1 is 300 ml and V2 is your unknown. Using this calculation for other questions be sure that units are all the same. So all molarities and all mL in this example.
You need 2,4 g NaOH (0,06 moles).
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
4 moles or 160 g NaOH is required for one litre solution.
Simple equality will do here as you have three things and require a fourth. (X L)(2.5 M NaOH) = (2.0 L)(0.75 M NaOH) 2.5X = 1.5 X = 0.60 Liters needed ================
to prepare 1N we have to dilute 40gms of NaOH in 1 litre of water as for NaOH normality =molarity so to prepare 0.1N NaOH we have to dilute 4gms of NaOH in 1 litre of water..
You need 2,4 g NaOH (0,06 moles).
Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.
4 moles or 160 g NaOH is required for one litre solution.
Simple equality will do here as you have three things and require a fourth. (X L)(2.5 M NaOH) = (2.0 L)(0.75 M NaOH) 2.5X = 1.5 X = 0.60 Liters needed ================
to prepare 1N we have to dilute 40gms of NaOH in 1 litre of water as for NaOH normality =molarity so to prepare 0.1N NaOH we have to dilute 4gms of NaOH in 1 litre of water..
300g NaOH + 700g water
0,4 g NaOH + 1000 ml H2O
First, find the molecular weights of the individual elements. Next, multiply each of the weights of the individual elements in the NaOH. Next, sum the molecular weights. What does the resulting number give you? That is, what does it express? Does it provide one of the entities needed to determine molarity? Next, determine how many 250-ml volumes are in a liter. You now have the two entities necessary to calculate the answer to your question. Hop to it.
no-AH
112,64 g of NaOH and 138,1 g of H2SO4
molarity equals moles of solute /volume of solution in litres . moles of NaOH equals 5g/40g = 0.125 and volume of solution will be volume of water + volume of NaOH = 0.5 litre+0.002 l which is nearly 0.5 litre . (volume of NaOH is calculated by its density) so molarity = 0.125mol/0.5litre = 0.25 M
"Dilute NaOH" without any other specifications in a chemistry lab generally refers to a 6M solution of NaOH in water.