One mole of NaOH weighs 23(Na)+16(O)+1(H)=40 g
A 0.05 M NaOH is then 2 g NaOH/L.
So dissolve 2 g NaOH in 500 ml water let it cool down to room temperature and fill it up to one liter. For accurate results you can use an analytical balance and a volumetric flask.
(40 g of NaOH dissolved in distilled water and make upto 1 litre gives 1 N)
DISSOLVE THE 40g in 1000ml of water it will give 0.1M NAOH.
Take 40 grams of NaOH in a beaker and dissolved in 1000ml (1litre) of distilled water.
5 M NaOH can be prepared by taking five moles of NaOH in 1000ml water.
Take bottle of 0.5 M NaOH.
Get auto-pipette calibrated to 50 μL.
Attach 50-250 μL tip.
Draw up NaOH.
Voilà. 50 μL of 0.5 M NaOH.
Dissolve 56.1g of KOH pellets in 1 dm3 of water.
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
no-AH
"Dilute NaOH" without any other specifications in a chemistry lab generally refers to a 6M solution of NaOH in water.
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
In order to prepare 50mM TES buffer, you will need to add in approximately 1000 ml of Proteinase K solution. From there, you will need to separate and stack the gels.
0,4 g NaOH + 1000 ml H2O
no-AH
"Dilute NaOH" without any other specifications in a chemistry lab generally refers to a 6M solution of NaOH in water.
You need 2,4 g NaOH (0,06 moles).
40 grams, this is the 1M NaOH standard laboratory solution.
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.
Hi, 6N NaOH = 6M NaOH 6M NaOH are 6 moles in 1L Mw (NaOH) = 39.88 gr/mole so: m = n x MW = 6 x 39.88 = 239.28 gr NaOH. :)
we need 0.8gm NaoH and dissolved in 10 ml of water to make 2N solution of NaoH .
Dissolve 0.4 g of NaOH in 100 ml of water. Try it out. Actually it is not suitable to prepare NaOH solutions in standard flasks.It should be made in beakers & must be standardised..This is done to find the correct normality...