Question

Question 1 (1 point) A 52.30 g sample of Ba(OH)2 is dissolved in enough water to make 1.20 L of solution. How many milliliters of this solution must be diluted with water in order to make 1.00 L of 0.100 M Ba(OH)2? 62.8 mL 254 mL O 305 ml O 523 mL O 393 mL O Question 2 (1 point) A chemist needs 225 mL of 2.3 M HCl. What volume of 12 M HCl must be dissolved in water to form this solution? (Ignore significant figures for this problem.) 12 mL 14 mL 8.2 mL O 23 mL O 43 mL O O

Answer 1

num Vou -0.254314 mi - Answer: 393mL 0.255 Ondar mass of Bath = 171.34 g/mol mass of Bapho used = 52.30g amber of moles of Boloth used mass 52:30g = 0.30524 moles one a saution - 1.201 molae mass 171.348/md . lity of BabHly solution Number of modes. 0.30524 nuts moovity a BabH)2 sduti on = 0.255m volume of solution 1.201 - This problem can be calculated by vmi = Vama. v = volume of Balth = To be calculated, m,-ndacity of BOOH 0.2350 V2 - volume of 0.im Ba@Hb = 1:00L; M - molocity 9 Ba@HE = 0.100m. ve Vama = 1:00LX o-loon _ 0.3921 L @ 393 mL 0.255 nb & vi- volume of 12 m Hd = x ; mi =ndacity of 12 m Hcl = 12.M . Va = volume of 2.3 m Hd = 225mL ; my=nolarity of 23 Hcl = 23 m2 volume of Hel must be dissolved (n)= va me 225 mLx2.30h - 43.12mL Ansurer : 45 mL 12 m ③ mass of 70% 0 HNo a solution - 103.89 Actual mass Hnos in Hnog solution - 103:89 * 10 = 72.66g , notae mass of HNO 8= 63.01g/mol Number of moles of Hnag= mass - 72669 = 1.153 moles - ndae mass 63.01g/md Anscurce 1015 md mass of solution = 167.9g, mass of solvent = 125.29 mass of solute = 167.99-125-29= 42.79 mass percentage: masse sdute X100 = 4279 X100 - 25.43% Answel 25.4% mass of solution 167.99 mi 100

6 A mixture of sand and water is Here sand does not dissolve in watee: So It is neither solution o aqueous solution so Ansice is none of the above- o cohen sauer saturated solution is added by a crystal. crystali zation takes place so answer is super saturated o volume of Nall solution = 65,0ml @ 0.065 L molacity of Nabb solution - 2.00m molar mass of Naßb= 102.894 g/mol molacity= mass of Nabh x 1000 mL, mass of Nabb-molauityxnolar nolae mass of Naßl volume massx volume 10oom mass of NaBb- 2.00 mote/jx 102.8949/ mdx 65.0mb to = 13.3769 Ansule: 13.4g 1000ml Nazpou will conizes as 3 Na and pow. so i mole of Napol Ionizes to 4 Ions. molacity of solution 0.60om; volume of solution - 999 [email protected] number of moles of Napou = molarity x vaume = 0.600 moel xo qaay Number of miles of Nag Pow=0.5994 Total number of moes of Na Pol = 0.5994* 4 = 2.976 moes. Ansuree : 2.40 moles @ mass of na, sou = 6:339; ndae mass = 142:049/mol volume of solution: 225 mL molocity - mass x 1000 har 6.339 x 1000 0-19806m I more mass volumem L 142.04ghroh 225m2 -25ML Answe o.198